Conditioning the ant to never pass by vertex $5$ is equivalent to cancelling edges $\{4,5\}$, $\{6,5\}$ and $\{8,6\}$. This leaves $7$ vertices and $9$ edges. The symmetry with respect to the plane containing $1$, $2$, $5$ and $8$ shows that $3$ and $7$ play the same role for a path starting from $1$ and ending at $8$. Likewise, $4$ and $6$ play the same role. So one can consider $3$ and $7$ as a single vertex, and $4$ and $6$ as a single vertex, only with more edges leading to them and leaving them than the others.
Thus the setting is equivalent to a random walk on a weighted graph with $5$ vertices and $5$ edges, drawn like a square plus an additional edge added to a vertex of the square. The square has vertices $1$, $2$, $3$ (which represents $3$ and $7$) and $4$ (which represents $4$ and $6$), the fifth vertex is $8$, and the edges are those of the square $\{1,2\}$, $\{2,3\}$, $\{3,4\}$, $\{4,1\}$, plus the additional edge $\{3,8\}$. Every edge has weight $2$ except the edge $\{1,2\}$ which has weight $1$. Weights on edges mean for instance that starting from $1$, one has $1/(1+2)$ chances to go to $2$ and $2/(1+2)$ chances to go to $4$.
Writing $t_i$ for the mean hitting time of $8$ starting from $i$ on the reduced graph, one looks for $t_1$. The vector $(t_1,t_2,t_3,t_4)$ solves the system $t_1=1+(t_2+2t_4)/3$,
$t_2=1+(t_1+2t_3)/3$, $t_3=1+(t_1+t_2+0)/3$ and $t_4=1+(t_1+t_3)/2$. Hence $t_1=62/5$ or something like that.
Alternatively, one can solve an analogous system on the original graph with $7$ vertices and $9$ edges. This system has size $6$ since $t_8=0$ and if one solves it one will note that some coordinates of the solution are equal, namely $t_3=t_7$ and $t_4=t_6$. The reduction explained above uses this fact to lower a priori the size of the linear system down to $4$ but it is not essential.
All this, and much more, in the must-read short book Random walks and electric networks by Peter G. Doyle and J. Laurie Snell.
An exact computation to answer question 1. is simple in principle, difficult to do exactly with no computer, and the result is frankly not very interesting. Question 2. is more interesting since, after 3600 steps on a graph of 25 vertices with diameter 10, the random walk is close to its stationary distribution.
The random walk is equivalent to the simple reversible random walk on the augmented graph where one adds 3 supplementary loops to each corner, 2 to each side vertex not a corner, and 1 to each other vertex. Every vertex in the augmented graph has degree 5 hence the stationary distribution is uniform. The Markov chain is aperiodic hence, at any large time, the probability to be at any of the vertices of this graph with 25 vertices is approximately 1/25.
The probability to be in any set S of vertices is approximately the size of S divided by 25. There are 4 corners hence the probability to be at one corner is approximately 4/25. There are 16 side squares hence the probability to be on one side is approximately 16/25.
Best Answer
Hint: We can draw a states diagram. Let $p_n$ be the probability that the ant reaches $n=5$ before $n=0$ starting at $n.$
First note that $p_5 = 1$ and $p_0 = 0.$Let's look at $p_4.$ There's a $\frac{1}{3}$ chance we go to $p_3$ and $\frac{2}{3}$ chance we go to $p_5.$ So $$p_4 = \frac{1}{3}p_3 + \frac{2}{3}p_5 = \frac{1}{3}p_3+\frac{2}{3}.$$
We write these equations for $p_1$ through $p_4.$ Then, solve for $p_3.$